To solve this problem, we can use the properties of the normal distribution, specifically the z-score formula.
Given that:
10% of the yarn faults are under 4 mm in length, this corresponds to the area to the left of 4mm.
90% of the yarn faults are under 20 mm in length; this corresponds to the area to the left of 20mm.
The normal distribution is symmetric, so we can find the z-score corresponding to the area to the left of 4mm and the z-score corresponding to the area to the left of 20mm.
First, we’ll find the z-scores for the 10th and 90th percentiles using a standard normal distribution table or a calculator:
For the 10th percentile (left of 4mm):
- Z10 = -1.2816
For the 90th percentile (left of 20mm):
- Z90 = 1.2816
The formula for the z-score is:
- Z = (X-μ)/σ
Where:
- X is the value (in this case, lengths 4mm and 20mm),
- μ is the mean of the distribution,
- σ is the standard deviation of the distribution.
From the given data, we know that the mean is the midpoint between 4mm and 20mm, which is
(4 + 20)/2 = 12 mm.
We can also find this using the z-score formula.
Using the z-score formula, we will get two equations as
For the 10th percentile (left of 4mm):
-1.2816 = (4 – μ )/σ
μ – 1.2816σ= 4 …………….1
For the 90th percentile (left of 20mm):
1.2816 = (20 – μ )/σ
μ +1.2816σ= 20 …………….2
Solving these equations will give us the standard deviation (σ) = 6.24 mm (Approx.) and Mean ( μ = 12 mm).
Therefore, the mean is 12 mm, and the standard deviation is approximately 6.24 mm.